Question

If ${\tan }^{-1}\left(\frac{a}{x}\right)+{\tan }^{-1}\left(\frac{b}{x}\right)=\frac{\pi }{2}$, then $x$ is equal to :

• A. $ab$
• B. $\frac{a}{b}$
• C. $\frac{b}{a}$
• D. $\sqrt{ab}$

Answer 1

The correct option is D $\sqrt{ab}$

${\tan }^{-1}\left(\frac{a}{x}\right)+{\tan }^{-1}\left(\frac{b}{x}\right)=\frac{\pi }{2}$

$\Rightarrow {\tan }^{-1}\left(\frac{a}{x}\right)=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{b}{x}\right)={\cot }^{-1}\left(\frac{b}{x}\right)$

$(\because {\tan }^{-1}\theta +{\cot }^{-1}\theta =\frac{\pi }{2})$

$\Rightarrow {\tan }^{-1}\left(\frac{a}{x}\right)={\cot }^{-1}\left(\frac{b}{x}\right)={\tan }^{-1}\left(\frac{x}{b}\right)$

$\Rightarrow \frac{a}{x}=\frac{x}{b}\Rightarrow x^2=ab$

$\Rightarrow x=\pm \sqrt{ab}$